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The following formula is given in our lab manual: $$ T = 2 \pi \sqrt{\frac{L}{g}} \left( 1 + \frac{1}{4}\sin^2 \frac{\theta}{2} + \frac{9}{64}\sin^4 \frac{\theta}{2}+\cdots \right) $$ for the period of a simple pendulum with length $L$ under the influence of gravitational acceleration with magnitude $g$. We ignore friction. I understand the solution of Newton's Equation via the small angle approximation $\sin \theta \approxeq \theta$ and why that gives the usual elementary formula $T = 2 \pi \sqrt{\frac{L}{g}}$. What I don't understand is how one derives the above formula for higher-order corrections to the period. I think the $\theta$ in the formula is the maximum angle to which the pendulum in question is swinging.

Any insight into the origin or derivation of this formula is appreciated.


marked as duplicate by Bill N, John Rennie newtonian-mechanics Jan 23 at 6:16

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    $\begingroup$ See the section “Legendre polynomial solution for the elliptic integral” at en.wikipedia.org/wiki/… $\endgroup$ – G. Smith Jan 22 at 17:24
  • $\begingroup$ BTW, there's a much nicer formula if you're doing this on a computer, using the AGM, the arithmetic-geometric mean, which converges very quickly. $\endgroup$ – PM 2Ring Jan 22 at 17:31
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    $\begingroup$ Possible duplicate of Period of a pendulum See the answer here. The solution is found in different textbooks, too. e.g., Engineering Dynamics by Beer & Johnston, , Classical Dynamics by Marion, This derivation is also a problem in Classical Mchanics by Taylor, so OP might be looking for that .... $\endgroup$ – Bill N Jan 22 at 18:30
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    $\begingroup$ That answer doesn’t explain how to get the series. $\endgroup$ – G. Smith Jan 22 at 18:43
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    $\begingroup$ @BillN I don't spend much time over here in Physics, but that seems pretty far from a direct duplicate. If I was searching for the series formula for the period I don't think that Q and A would be directly helpful. $\endgroup$ – James S. Cook Jan 23 at 1:14

If we take the zero of gravitational potential energy to be at the pivot, then by conservation of energy,

$$\frac{1}{2}m(L\dot{\theta})^2-m g L\cos{\theta}=-m g L\cos{\theta_0}$$

where $\theta_0$ is the maximum angle (where the kinetic energy is zero). This gives the differential equation


which gives the period as four times the time to go from 0 to $\theta_0$:


Using the trigonometric identity


this becomes


With the change of integration variable to $u$ where


this becomes




Now use a Taylor expansion and integrate term-by-term to get the result you were given:

$$\begin{align} T&=4\sqrt\frac{L}{g}\int_0^{\pi/2}du\;\left(1+\frac{1}{2}k^2\sin^2{u}+\frac{3}{8}k^4\sin^4{u}+...\right)\\ &=4\sqrt\frac{L}{g}\left(\frac{\pi}{2}+\frac{1}{2}k^2\frac{\pi}{4}+\frac{3}{8}k^4\frac{3\pi}{16}+...\right) \\ &=2\pi\sqrt\frac{L}{g}\left(1+\frac{1}{4}k^2+\frac{9}{64}k^4+...\right). \end{align}$$

  • $\begingroup$ I think you'd enjoy The AGM Simple Pendulum by Mark B. Villarino. $\endgroup$ – PM 2Ring Jan 22 at 18:45
  • $\begingroup$ Thanks for this answer. Exactly the sort of detail I had hoped for. $\endgroup$ – James S. Cook Jan 22 at 21:21

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